Respuesta :
Answer:
0.18173219.
Step-by-step explanation:
We have been asked to find what will be the probability that at least 6 of 9 adults use their smartphones in meetings or classes.
We will find our answer using Bernoulli's trails.
[tex]_{r}^{n}\textrm{c}\cdot p^{r}\cdot (1-p)^{n-r}[/tex]
First of all we will find the probabilities when r is 6, 7,8 and 9 then we will add them all.
When r=6,
[tex]_{6}^{9}\textrm{c}\cdot 0.46^{6}\cdot (1-0.46)^{9-6}[/tex]
[tex]\frac{9!}{6!3!} *0.46^{6} *0.54^{3}[/tex]
[tex]\frac{9*8*7*6!}{6!*3*2*1} *0.009474296896*0.157464[/tex]
[tex]84*0.009474296896*0.157464=0.12532[/tex]
Similarly we will find Probabilities when r=7, 8 and 9.
When r=7
[tex]_{7}^{9}\textrm{c}\cdot 0.46^{7}\cdot (1-0.46)^{9-7}[/tex]
[tex]\frac{9!}{7!2!} *0.46^{7} *0.54^{2}[/tex]
[tex]\frac{9*8*7!}{7!*2*1} *0.00435817657216*0.2916[/tex]
[tex]36*0.00435817657216*0.2916=0.04575039[/tex]
When r=8
[tex]_{8}^{9}\textrm{c}\cdot 0.46^{8}\cdot (1-0.46)^{9-8}[/tex]
[tex]\frac{9!}{8!1!} *0.46^{8} *0.54[/tex]
[tex]\frac{9*8!}{8!*1!} *0.00200476*0.54[/tex]
[tex]9 *0.00200476*0.54=0.0097431336[/tex]
When r=9,
[tex]_{9}^{9}\textrm{c}\cdot 0.46^{9}\cdot (1-0.46)^{9-9}[/tex]
[tex]\frac{9!}{9!0!} *0.46^{9} *1[/tex]
[tex]1 *0.00092219*1=0.00092219[/tex]
Now let us add all the probabilities to get the final answer.
[tex]0.12532+0.04575+0.00974+0.00092219=0.18173219[/tex]
Therefore, probability that at least 6 of 9 adults use their smartphones in meetings or classes is 0.18173219.
The probability that at least [tex]6[/tex] out of [tex]9[/tex] adult use their smartphone in meeting or classes is [tex]\fbox{\begin\\\ 0.1817\\\end{minispace}}[/tex].
Further explanation:
It is given that [tex]46\%[/tex] smartphone user use their smartphone in meetings or classes and at least [tex]6[/tex] out of [tex]9[/tex] adult smartphone user are selected randomly.
Here we will use the concept of Binomial probability.
If an experiment is performed [tex]n[/tex] times and it has only two outcomes that is, “success” and “failure”.So, the probability associated with this experiment is called Binomial probability.
The probability to get exactly [tex]r[/tex] successes in [tex]n[/tex] trial is given as follows,
[tex]\boxed{P=^{n}C_{r}p^{r}(1-p)^{n-r}}[/tex] ......(1)
Here, [tex]P[/tex] is the probability of [tex]r[/tex] successes in [tex]n[/tex] trials.
About [tex]46\%[/tex] smartphone user use their smartphone in meetings or classes that means the probability of the smartphone user use their smartphone in meetings or classes is as follows:
[tex]\begin{aligned}46\%&=\dfrac{46}{100}\\&=0.46\end{aligned}[/tex]
The statement, “at least six smartphone user” means six or more smartphone user. We will calculate the probability of the [tex]6,7,8[/tex] and [tex]9[/tex] adult smartphone user.
Substitute [tex]0.46[/tex] for [tex]p[/tex], [tex]9[/tex] for [tex]n[/tex] and [tex]6,7,8,9[/tex] for [tex]r[/tex] in equation (1) to obtain the probability of summation as follows,
[tex]P=^{9}C_{6}(0.46)^{6}(1-0.46)^{9-6}+^{9}C_{7}(0.46)^{7}(1-0.46)^{9-7}+^{9}C_{8}(0.46)^{8}(1-0.46)^{9-8}+\qquad^{9}C_{9}(0.46)^{9}(1-0.46)^{9-9}\\\\P=\dfrac{9!}{6!\cdot(9!-6!)}\cdot(0.46)^{6}\cdot(0.54)^{3}+\dfrac{9!}{7!\cdot(9!-7!)}\cdot(0.46)^{7}\cdot(0.54)^{2}+\dfrac{9!}{8!\cdot(9!-8!)}\cdot(0.46)^{8}\cdot(0.54)^{1}+\dfrac{9!}{9!\cdot(9!-9!)}\cdot(0.46)^{9}\cdot(0.54)^{0}[/tex]
Further solving the above equation as follows,
[tex]\begin{aligned}P&=84\cdot(0.46)^{6}\cdot(0.54)^{4}+36\cdot(0.46)^{7}\cdot(0.54)^{2}+9\cdot(0.46)^{8}\cdot(0.54)^{1}+1\cdot(0.46)^{9}\cdot(0.54)^{0}\\&=0.1253+0.04575+0.009743+0.00092219\\&=0.1817\end{aligned}[/tex]
Therefore, the probability that at least [tex]6[/tex] out of [tex]9[/tex] adult use their smartphone in meeting or classes is [tex]\boxed{0.1817}[/tex].
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Answer details:
Grade: College
Subject: Mathematics
Chapter: Probability
Keywords: Probability, numbers, chances, smartphone, meeting, classes, 0.1817, summation, user, equation, selected, Binomial probability, randomly.
